A charge of 5 nC is uniformly distributed on a ring of radius 4 cm. Find the potential at the axis at a distance of 3 cm from the center.

Options

A. `0.09 V`
B. 600 V
C. 900 V
D. 300 V

Answer

To find the electric potential at a point on the axis of a uniformly charged ring, we can use the formula for the potential \( V \) at a distance \( x \) from the center of the ring:

\[ V = \frac{kQ}{\sqrt{x^2 + a^2}} \]

where:
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( Q \) is the total charge on the ring,
- \( x \) is the distance from the center of the ring along the axis,
- \( a \) is the radius of the ring.

### Step-by-Step Solution:

1. **Identify the given values**:
- Charge \( Q = 5 \, \text{nC} = 5 \times 10^{-9} \, \text{C} \)
- Radius of the ring \( a = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m} \)
- Distance from the center along the axis \( x = 3 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \)

2. **Substitute the values into the formula**:
\[ V = \frac{9 \times 10^9 \times 5 \times 10^{-9}}{\sqrt{(3 \times 10^{-2})^2 + (4 \times 10^{-2})^2}} \]

3. **Calculate the denominator**:
- First, calculate \( (3 \times 10^{-2})^2 \):
\[ (3 \times 10^{-2})^2 = 9 \times 10^{-4} \]
- Next, calculate \( (4 \times 10^{-2})^2 \):
\[ (4 \times 10^{-2})^2 = 16 \times 10^{-4} \]
- Now, add these two results:
\[ 9 \times 10^{-4} + 16 \times 10^{-4} = 25 \times 10^{-4} \]
- Finally, take the square root:
\[ \sqrt{25 \times 10^{-4}} = 5 \times 10^{-2} \]

4. **Substitute back into the potential formula**:
\[ V = \frac{9 \times 10^9 \times 5 \times 10^{-9}}{5 \times 10^{-2}} \]

5. **Simplify the expression**:
- The \( 5 \) in the numerator and denominator cancels out:
\[ V = \frac{9 \times 10^9}{10^{-2}} = 9 \times 10^9 \times 10^2 = 9 \times 10^{11} \]

6. **Final calculation**:
\[ V = 900 \, \text{V} \]

### Final Answer:
The potential at the axis at a distance of 3 cm from the center of the ring is **900 volts**.